Physics - Vectors Question with Solution | TestHub

According to the question, $A+B=C\dots ..(0)$ Also, when $B$ is doubled, $C$ becomes perpendicular to $A$. Let the new magnitude of $C$ be $C^{\prime }$ That means $A.C^{\prime }=0\left(A{C}^{\prime }\cos 90=0\right)$ $>(A)(A+2B)=0\dots \dots .(1)$ $>\left(A^{\wedge }2+2AB\right)=0$ $>|A|=-2AB$ To find the magnitude of $C$, square eq. (0) $>(A+B{)}^{\wedge }2=|C|$ $>|A|+|B|+2AB$ but $2AB=-|A|.$ from $(1)$ which means $|C|=|B|$